# Probability - Independence - boxes and balls

$Box A$: 2 gray balls 1 white ball $Box B$: 1 gray ball 3 white balls. We choose randomly a box and choose randomly one ball. (1) Calculate the probability that the ball is white. My attempt:$0.5*(1/3)+0.5*(3/4)=13/24$ (2) Calculate the probability that Box A was chosen when it is known that white ball was chosen. My attempt: $[0.5*(1/3)]/(first question)$ (3) We return the ball to the box from which it was taken and now we take randomly another ball . calculate the probability that the other ball is white when it is known that the first one is white. My attempt: $(first question)*(first question)/(first question)=13/24$(4) Like in the third question except now the first ball stays out. My attempt: ${0.5*(1/3)+0.5*0.5+0.5*(3/4)}/(first question)$

Given that the chosen ball was white, the probability it came from A is $p_A=\frac{(0.5)(1/3)}{13/24}$, and the probability it came from B is $p_B=\frac{(0.5)(3/4)}{13/24}$.
Thus the (conditional) probability the next ball is white is $(1/3)p_A+(3/4)p_B$.
For d), the $1/3$ is replaced by $0/2$ and the $3/4$ is replaced by $2/3$.